sum of binomial coefficients expansion to prove equation I want to prove that
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \binom{n+1}{3}
$$
I already expanded
$$
\binom{n+1}{3}
$$
to
$$
\binom{n+1}{3} = \frac{1}{6} * (n+1) *n*(n-1)
$$
and I know that the following equation must be right
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \frac{1}{6} * (n+1) *n*(n-1)
$$
but I do not get the expansion right, I tried starting with writing the sum explicit
$$
\binom{1}{2}+\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
since 1 over 2 is zero it can be shortened to
$$
\sum_{i=2}^{n}{\binom{i}{2}} = \binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
then I expanded the binomial coefficients to the corresponding factorial form
$$
\binom{n}{k} = \frac{n!}{k! * (n-k)!}
$$
but I do not get it right, could someone please help me?
 A: $$\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}=\frac{1\times 2}{2}+\frac{2\times 3}{2}+\frac{3\times 4}{2}+...+\frac{(n-1)\times n}{2}=\frac{1}{2}(1\times 2+2\times 3+3\times 4+...+(n-1)\times n)=\frac {1}{2}\times \frac{(n-1)n(n+1)}{3}=\frac{(n-1)n(n+1)}{6}$$
A: $\begin{array}\\
\binom{n+1}{3}-\binom{n}{3}
&=\dfrac{(n+1)n(n-1)}{6}-\dfrac{n(n-1)(n-2)}{6}\\
&=\dfrac{n(n-1)((n+1)-(n-2))}{6}\\
&=\dfrac{3n(n-1)}{6}\\
&=\dfrac{n(n-1)}{2}\\
&=\binom{n}{2}\\
\end{array}
$
Therefore
$\sum_{i=1}^{n}\binom{i}{2}
=\sum_{i=1}^{n}(\binom{i+1}{3}-\binom{i}{3})
=\binom{n+1}{3}
$.
Note that
$\binom{n}{m} = 0$
for $m > n$.
In general,
since
$\binom{n}{m}
=\dfrac{\prod_{k=0}^{m-1}(n-k)}{m!}
$,
$\begin{array}\\
\binom{n+1}{m}-\binom{n}{m}
&=\dfrac{\prod_{k=0}^{m-1}(n+1-k)}{m!}-\dfrac{\prod_{k=0}^{m-1}(n-k)}{m!}\\
&=\dfrac{\prod_{k=-1}^{m-2}(n-k)}{m!}-\dfrac{\prod_{k=0}^{m-1}(n-k)}{m!}\\
&=\dfrac{(n+1)\prod_{k=0}^{m-2}(n-k)-(n-m+1)\prod_{k=0}^{m-2}(n-k)}{m!}\\
&=\dfrac{((n+1)-(n-m+1))\prod_{k=0}^{m-2}(n-k)}{m!}\\
&=\dfrac{m\prod_{k=0}^{m-2}(n-k)}{m!}\\
&=\dfrac{\prod_{k=0}^{m-2}(n-k)}{(m-1)!}\\
&=\binom{n}{m-1}\\
\end{array}
$
so that
$\sum_{i=1}^{n}\binom{i}{m-1}
=\sum_{i=1}^{n}(\binom{i+1}{m}-\binom{i}{m})
=\binom{n+1}{m}
$.
A: A variation based upon the  binomial theorem and the finite geometric series formula.
We consider the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. We can write this way
\begin{align*}
\binom{n}{k}=[x^k](1+x)^n\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{i=1}^n\binom{i}{2}}&=\sum_{i=0}^n[x^2](1+x)^i\tag{2}\\
&=[x^2]\sum_{i=0}^n(1+x)^i\tag{3}\\
&=[x^2]\frac{(1+x)^{n+1}-1}{(1+x)-1}\tag{4}\\
&=[x^3]\left((1+x)^{n+1}-1\right)\tag{5}\\
&\,\,\color{blue}{=\binom{n+1}{3}}\tag{6}
\end{align*}
and the claim follows.

Comment:

*

*In (2) we apply the coefficient of operator according to (1). The lower limit of the index $i$ is set to $i=0$ which doesn't change anything, since $\binom{0}{2}=[x^2](1+x)^0=[x^2]1=0$.


*In (3) we use the linearity of the coefficient of operator: $[x^p]A(x)+[x^p]B(x)=[x^p]\left(A(x)+B(x)\right)$.


*In (4) we use the finite geometric summation formula.


*In (5) we observe the denominator is $x$ and we use the rule $[x^p]\frac{1}{x}A(x)=[x^{p+1}]A(x)$.


*In (6) we select the coefficient of $x^3$.
